Help with Area of parametric equations problem

Mcbrown108
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Homework Statement



Find the area of the region enclosed by the parametric equation
x=t^3-8t
y=2t^2





The Attempt at a Solution


I am not even sure how to start this problem.
I read somewhere that to start with you solve for t in one of the equations.
when i solve for t I end up with really weird equations.

Can anyone help?
 
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There must be more information, does it say how the region is enclosed? I'm assuming it's with the x axis. Always be sure to include all the information in the question.
 
That is all that is given to me. I thought I was missing some information as well.
 
Oh, that's really, really wierd! I was all set to agree with gamesguru but, just to make sure, I graphed it on my TI-83. That is a closed curve with t going from -\sqrt{8} to \sqrt{8}.

You should be able to find the area by using the fact that
\int y(x)dx= \int y(t)\frac{dx}{dt}dt
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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